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Curvature of Plane Graphs

Learn to derive and use the curvature formula for a graph, and see how its slope and second derivative affect curvature at a point.

Intermediate 9 min read

What You'll Learn

  • Represent a graph as a vector-valued curve and apply the cross-product curvature formula
  • Derive the graph formula from the curve’s velocity and acceleration
  • Calculate curvature at specified points, showing each derivative and substitution
  • Identify when curvature equals the absolute value of the second derivative
  • Avoid confusing a graph’s second derivative with its curvature

Curvature of a Plane Graph

A graph can bend sharply in one region and gently in another. Curvature gives a numerical measure of this bending at a point. For a graph \(y=f(x)\), the second derivative is important, but it is not by itself the curvature in general. The graph’s slope also affects the answer.

In this tutorial, we derive the formula for curvature of a plane graph from the cross-product formula for a regular space curve. We then use it to calculate curvature at specific points and interpret the roles of the first and second derivatives.

Key formula: If \(f\) is twice differentiable near \(x=a\), then the graph \(y=f(x)\) has curvature at \(x=a\) \[ \kappa(a)=\frac{|f''(a)|}{\bigl(1+[f'(a)]^2\bigr)^{3/2}}. \]

Representing a Graph as a Curve

To apply a curve formula to the graph \(y=f(x)\), use \(x\) as the parameter. In three-dimensional coordinates, the plane graph can be represented by

$$ \mathbf{r}(x)=\langle x,f(x),0\rangle. $$

The curve is regular wherever it is defined: its velocity has first component \(1\), so it cannot be the zero vector. If \(f\) is twice differentiable, the curve has the derivatives needed for the cross-product curvature formula. That formula states that, for a twice-differentiable regular curve \(\mathbf{r}\), its curvature is

$$ \kappa(x)=\frac{\left|\mathbf{r}'(x)\times\mathbf{r}''(x)\right|}{\left|\mathbf{r}'(x)\right|^3}. $$

Here, \(|\mathbf{v}|\) denotes the magnitude of a vector \(\mathbf{v}\). The numerator measures the magnitude of the cross product of the velocity and acceleration, while the denominator accounts for the speed of the parametrization. We now calculate both quantities for a graph.

Deriving the Graph Formula

Differentiate the graph’s position vector component by component. Its first derivative is the velocity, and its second derivative is the acceleration:

$$ \mathbf{r}'(x)=\langle 1,f'(x),0\rangle, \qquad \mathbf{r}''(x)=\langle 0,f''(x),0\rangle. $$

The speed is the magnitude of the velocity. The cross product can be calculated from the determinant of the component vectors:

$$ \left|\mathbf{r}'(x)\right| =\sqrt{1^2+[f'(x)]^2+0^2} =\sqrt{1+[f'(x)]^2}, $$
$$ \mathbf{r}'(x)\times\mathbf{r}''(x) = \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 1&f'(x)&0\\ 0&f''(x)&0 \end{vmatrix} =\langle 0,0,f''(x)\rangle. $$

Therefore, the magnitude of the cross product is \(|f''(x)|\). The cube of the speed is \(\bigl(1+[f'(x)]^2\bigr)^{3/2}\). Substituting these two calculations into the cross-product curvature formula gives the graph formula.

Derivation: For a twice-differentiable function \(f\), the graph curve \(\mathbf{r}(x)=\langle x,f(x),0\rangle\) is regular. Its velocity has magnitude \(\sqrt{1+[f'(x)]^2}\), and the magnitude of its velocity–acceleration cross product is \(|f''(x)|\). Thus \[ \kappa(x)=\frac{|f''(x)|}{\bigl(1+[f'(x)]^2\bigr)^{3/2}}. \]

This derivation also explains why the absolute value appears: curvature measures the magnitude of bending, not whether the graph bends upward or downward. The sign of \(f''(x)\) distinguishes those directions, but curvature itself is nonnegative.

Applying the Formula at a Point

To find the curvature at \(x=a\), first calculate \(f'(a)\) and \(f''(a)\). Then substitute those values into the formula. The point on the graph is \((a,f(a))\); the value of \(f(a)\) identifies the location, while the curvature calculation uses the first and second derivatives at \(a\).

Worked Example: Curvature of a Parabola

Find the curvature of \(y=x^2\) at \(x=1\).

First identify the point on the graph: \(f(1)=1^2=1\), so the point is \((1,1)\). Differentiate:

$$ f(x)=x^2,\qquad f'(x)=2x,\qquad f''(x)=2. $$

At \(x=1\), the slope is \(f'(1)=2(1)=2\), and the second derivative is \(f''(1)=2\). Substitute both values:

$$ \kappa(1) =\frac{|2|}{(1+2^2)^{3/2}} =\frac{2}{5^{3/2}} =\frac{2}{5\sqrt{5}}. $$

As a check, the graph curve has \(\mathbf{r}'(1)=\langle 1,2,0\rangle\) and \(\mathbf{r}''(1)=\langle 0,2,0\rangle\). Their cross product is \(\langle 0,0,2\rangle\), of magnitude \(2\), and the velocity has magnitude \(\sqrt{5}\). The cross-product formula gives \(2/(\sqrt{5})^3=2/(5\sqrt{5})\), matching the result.

Worked Example: Curvature with a Nonzero Slope

Find the curvature of \(y=x^3/3\) at \(x=1\).

The point is \((1,1/3)\), since \(f(1)=1/3\). The derivatives are

$$ f'(x)=x^2,\qquad f''(x)=2x. $$

Thus \(f'(1)=1\) and \(f''(1)=2\). The curvature is

$$ \kappa(1) =\frac{|2|}{(1+1^2)^{3/2}} =\frac{2}{2^{3/2}} =\frac{2}{2\sqrt{2}} =\frac{1}{\sqrt{2}}. $$

For a check using the curve derivatives, \(\mathbf{r}'(1)=\langle 1,1,0\rangle\) and \(\mathbf{r}''(1)=\langle 0,2,0\rangle\). Their cross product has magnitude \(2\), and the speed is \(\sqrt{2}\). Hence curvature is \(2/(\sqrt{2})^3=2/(2\sqrt{2})=1/\sqrt{2}\), as calculated.

Worked Example: A Straight Line with Nonzero Slope

Find the curvature of \(y=3x-2\) at \(x=4\).

The point is \((4,10)\), because \(f(4)=3(4)-2=10\). Differentiate:

$$ f'(x)=3,\qquad f''(x)=0. $$

At \(x=4\), the slope is \(3\) and the second derivative is \(0\). Therefore,

$$ \kappa(4) =\frac{|0|}{(1+3^2)^{3/2}} =\frac{0}{10^{3/2}} =0. $$

For a direct check, \(\mathbf{r}'(4)=\langle 1,3,0\rangle\) and \(\mathbf{r}''(4)=\langle 0,0,0\rangle\). Their cross product is zero, so the cross-product formula also gives zero curvature. The line has nonzero slope but does not bend.

How Slope and Second Derivative Affect Curvature

The numerator \(|f''(x)|\) reflects how quickly the graph’s slope changes as \(x\) changes. The denominator adjusts for the fact that curvature concerns bending with respect to distance along the curve, rather than simply change with respect to the horizontal coordinate. In particular, a graph with a steeper slope has a larger denominator when \(|f''(x)|\) is held fixed.

Since \(1+[f'(x)]^2\geq 1\), its power \(3/2\) is also at least \(1\). This leads to a precise comparison:

  • If \(f'(x)=0\), then \(\kappa(x)=|f''(x)|\).
  • If \(f''(x)=0\), then \(\kappa(x)=|f''(x)|=0\), regardless of the slope.
  • If \(f'(x)\ne 0\) and \(f''(x)\ne 0\), then the denominator is strictly greater than \(1\), so \(0\leq\kappa(x)<|f''(x)|\).

Thus, the numerator equals the curvature when the slope is zero or when the second derivative is zero. A nonzero slope makes curvature strictly smaller than \(|f''(x)|\) only when the second derivative is also nonzero.

A Common Mistake

A common error is to use only the second derivative and write \(\kappa(x)=|f''(x)|\) at every point. For \(f(x)=x^2\) at \(x=1\), this incorrect calculation gives

$$ \kappa(1)\ \text{(incorrect)} =|f''(1)| =|2| =2. $$

It omits the slope factor. Since \(f'(1)=2\), the denominator is \((1+2^2)^{3/2}=5^{3/2}\), not \(1\). The corrected calculation is

$$ \kappa(1) =\frac{|f''(1)|}{(1+[f'(1)]^2)^{3/2}} =\frac{2}{5^{3/2}}. $$

The special case \(f''(x)=0\) is worth checking when reasoning about comparisons. Even if the slope is nonzero, the curvature and \(|f''(x)|\) are then both zero; the strict inequality \(\kappa(x)<|f''(x)|\) does not hold.

Conditions and Interpretation

The formula applies at points where \(f\) is twice differentiable. The graph parametrization is regular wherever \(f\) is defined, because its velocity’s first component is \(1\). If \(f\) is not twice differentiable at a point, this formula does not establish that a classical curvature exists there; the point must be examined using an appropriate definition or method.

Curvature is always nonnegative. For example, \(f''(x)<0\) does not produce negative curvature: the absolute value in the numerator ensures that the formula measures bending magnitude. If the coordinates represent physical lengths, the slope is dimensionless, while \(f''\) and curvature have units of inverse length.

The formula is useful whenever a curve is naturally given as a graph and its derivatives are accessible. It makes the calculation at a specified point systematic: evaluate the slope and the second derivative there, preserve the absolute value, and include the full denominator. Neither the slope nor the second derivative alone determines curvature in general.

Check Your Understanding

Use the graph curvature formula to answer these questions.

  1. What condition ensures that the graph parametrization \(\mathbf{r}(x)=\langle x,f(x),0\rangle\) is regular?
  2. For \(f(x)=x^2\), what are \(f'(1)\) and \(f''(1)\), and what curvature formula results at \(x=1\)?
  3. Why does the curvature formula use \(|f''(x)|\) rather than \(f''(x)\)?
  4. If \(f'(a)\ne 0\) and \(f''(a)\ne 0\), how does \(\kappa(a)\) compare with \(|f''(a)|\)?
  5. Can a graph have nonzero slope and zero curvature at a point? Give an example.

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